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# 15th - 1 1.7 P)VVICR SERIES 1 11โ1 2 lnn Uโโโ< 1...

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Unformatted text preview: 1 1.7 P( )VVICR SERIES . _ 1; 11โ1 _ 2. lnn Uโโโ < 1 2:ยป lnn (x l < 1 5โ < 1 6 < X < 4', when X = โ(3 we have 11 โ1 oc- Us 11 โ1 oc- (X1 .1โ TX! 7:: z (โ1)โ, a divergent series; when X = โ4 we have 2: 1, a divergent series not 1171 (a) the radius is 1; the interval of convergence is โ6 < X < โ4 (b) the interval of absolute convergence is 76 < X < 74 (c) there are no values for which the series converges conditionally - 11.,+ - (xโ2]"_l 1on xโ2 5.1111111โ uโ"J<1=>nlll]}x__l W'W 2:ยป โ8 < X < 12; when X = โ8 we have (โ1)โ, a divergent series; when X = 12 we have 1, a divergent 11โ1 11โ] series (a) the radius is 10; the interval of convergence is โ3 < X < 12 (b) the interval of absolute convergence is โ3 < X < 12 (c) there are no values for which the series convelges conditionally 1) - uniโ - 10'โ 7 ntlnnli - r1 - lnn โ 27' llllpbc- 1.1.. <1 2:} 111โ19139 (n11](lntn1]1_โ1"โ K" < l 2:} lxl (nlโlpbc- n11) lntntlJ) < l 1. 3 2 =>|X|(1) lim_ (1โ) <1=>|X|(li1n_m1) <1=>|X|<1=>โ1<X<l;whenX=โlwehave [lโโCXJ 11โโ09 n TI: TI: (โ1 1" - _. ,. _ _. 1 - . E l wlnch conv erges absolutel) , when X โ 1 we ha\ e E m Wthll converges n_ nโl (a) the radius is 1; the interval of convergence is 71 < X < 1 (b) the interval of absolute convergence is โ1 < X < 1 (c) there are no values for which the series convelges conditionally 2n+3 xโ n 32. Hut โ1โโ <1 2:ยป li1n 145] 11โ100 Us 11โ100 ( I) 2 _ 2 . ::>(x.โ:/โ]<1::>(Xโ\,/ยง) <2::> Xโ\/ยง<\/ยง:ยข>โ 6<Xโ\/ยง<\/ยง:ยข>0<X<2\/ยง;when 211โ xโ โ1ยป 2 3ยฐ :1 1 2 3ยฐ _ , , . . X = Owe have 2: \$ = โ Z 2 gโ = โ Z we which diverges s1nce โ11111)C an \$ 0; when X = EVE nโ1 nโ1 nโl โ I 2n+l so ,5 so so \, . n 12 . . we have 2: ( 2]" = Z 3 gโ = 2: V6, a d1vergent series nโl nโl 11โ] (a) the radius is V5; the interval of convergence is 0 < .X < 2V5 (b) the interval of absolute convergence is 0 < X < 2V5 (c) there are no values for which the series convelges conditionally . ml 2 s - ., 37.nliโmxl โ:1: <1 => โ11ng (โ13โ) 1%)โ <1=>("3โ-โn1_11nw|1|<1=> *3โ <1 => x~<2 2:ยป < x/E 2:ยป โ G < .X < x/E; at X = ; VG we have (1)rl which diverges; the interval of convergence is nโCl x. _l n โ 6 < X < x/E ; the series 2: (k 3' 1) is a convelgent geometric series when โ\,/E < .X < V5 and its sum is 1-1โ0 11.8 TAYLOR AND MACIAURIN SERIES 3. ๏ฌx) 2 3? = f], fโ(x) = โx_2, f"(x) 2 211โ3, fโ"(:x) = โ6x_4; f(2) , fโ(2) = โ ET, fโ(2) = 1:, fโโ(x) = โ =>P.,(x)=%,P](x)=%โ}T(xโ2),P3(x)=%โ3T(xโ2)โ%(xโ2)', P;1(x)=?3 11x 2) 11x 213 3,11: 21" 6. f(x)=cosx,fโ(x)=โsinx,f"(x)=โcosx,fโโ(x)=sinx;f(ยง]=eosgz๏ฌ, PG] :โsin%:โ7]E,fโ(%] :โe05ยง:โ7]E,fโโ(%] :sin๏ฌz๏ฌ 2:ยป P1100: V12, P100=๏ฌg-๏ฌgb-๏ฌthth๏ฌ-๏ฌg๏ฌ-๏ฌ-๏ฌยง(x-%Ji 2 :1 ๏ฌatwareโ452%) โ,3โ51xโ11 12. ๏ฌx): (1 โx)_] => ๏ฌx): (1 โx)โ3.fโ(x) = 211โ 111โ3, 1%) = 3:11 โ 111โ4 => 1mm 2 ma โ 11-11-2110) 2 1, 1'10) 2 1, 9โ10) 2 2, 1111(0) = 3:, ,fmi-(m = k! '35 =1โxโx2โXโโ... =2: xโ :8 1โK n-D ๏ฌx) 2 3xโ โ x4 โ 2.313 โ x2 โ 2 2:ยป fโ(x) = 15.31โ1 โ 4x3 โ 6x2 โ 2x, f"(x) = 60113 โ 12.313 โ 12xโ 2, 1mm 2 180112 โ 24.x โ 12,1001โ: 3601: โ 24, ยฃ15101) 2 360, Mm = 0 ifn :-~ 6; f(โ 1) = โ7, 91โ11 = 23,1"1โ1): โ82,fโโ(โ1) = 216, 1(411โ1) = โ334, 1(5?(โ 1) = 360, 1031โ1) 2 Oifn :5 6 24. =1ยป 3.1โ x4 2113 x3 2 = 7 23(11 1) 3%: 1)2 231,601 11" 343,411: 1)4 355,001 1)โ = 7 231โ): 1) 41(11 1].2 36(x 1)?โ 1601 1)โ1 3(11 1)โ 11.9 CONVERGENCE OE TAYLOR SERIES; ERROR liS'lโIPB/IA'IES . L2 2" _ _ 1:211 ; 113 35 1โD" _ n31: 6- = 1111;: j (5โ,) = ((1%) ) = 2;] 1 = z 1']โ Flโ nโ : 1 _ _ x_โ _ x_"' _ 2-2I 21-4! 24-6! 15 If; : Kenโ121]: K3 Z;an : Zoznxnm : X3 _ 2x3. _23K4 _ 23.x". _ 22 V l โ x = 1 โ 5, โ g: โ โ . By the Alternating Series Estimation Theorem the |err01'| < % < @ = 1.25 X 10โโ 33. [5111'] 1: when x = %;the sum is tanโ1 :3 0.808448 11.10 APPLICA'IโIONS OE POWER SERIES ._--) . _1 3 1โ _L _2 _ยง':1" , 7- [l_xโ..,]1;-=l Fi๏ฌ ( .,](2!]1x) (ยฃ11 1;!( 111x) :1 13x1 gxh 156x: 13. (1 โ 211)3 = 1 31 2x) (MELโ2""! 13โโ131'2โ = 1 6x 12112 3114* 24. [1โ x3] 3/ โ 211)? = a] (2513 2am): (3613 251] 511))? (4:14 2:13 233))(3 (narl nan_g)xโ = 0 2:ยป a] = 0, 2:13 โ 2aโ 2 0, 333 โ 351] = 0, 4:14 โ 4513 = 0 and in general naJ1 โ nan_2 = 0. Since y = 3 when x: 0,we]1a\-'e aโ = 3. Therefore :13 = โ3,ay, 20,34 = 3, ,Elgm] 20,512rl = (โ1)โ3 ,_ 2 .1 _ 3ยฐ 2 _ 3ยฐ 2 n_ 3 2:ยป }โ3โ3x โ3x โ... โZ%)3(โl)โxโโZ%3[โx] โm_r A 3 _1_ 31. }โโx3y=233 633): (4-33.; am}? (n(n Dan an_4)xโโ2โ . 2x 2:ยป 233:0,63321, 4 - 334 โ an 2 0, 5 - 435 โ a] = 0, and in general n(n โ Dan โ an_4 = 0. Since 3/ = b and y = a when x = CI, wehavean : and a] = b. Therefore :12 = 0,02. = = โ ๏ฌns : โ 4%โ -. 20,617 = ifโ , _ 1 :1. a 4 b r. 1 7 2m" bx" 2* .โ โ a b" 2-3 X 3-4 x 4-5 X 2-3-07 9โ 3 4 s 4 5-3-9 F(X):J:([2_%_%_%_HI)dt:[In 1|] 1L:- ;. _ :ยป |e1'1'01'| < :5 0.000013 54.0: l)sin(x]}]]=(x 1)(X1H 1 1 .โ,)=1_ 1 20m 11:1 S!(x} 11-1 you 1:? _ sum 11' _ ::>xlip})cl(xโl)5in(x:]]: lim_l (1 1 1 .mjzl ...
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15th - 1 1.7 P)VVICR SERIES 1 11โ1 2 lnn Uโโโ< 1...

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